I have a real valued number yt. At each time step t, yt is multiplied by (1+ϵ) with probability p and multiplied by (1−ϵ) with probability 1−p. What is the expected value of yt+n? What is the variance?

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encolatgehu · Accepted Answer

It is quite simple if you use independence in a more direct way and the method works for any distribution. Assuming that yn=∏1≤k≤nXk where the Xk are i.i.d. factors. By independence:  Eyn=E∏kXk=∏kEXk=(EXk)n  valid for any distribution. In the specific (Bernoulli) example:   EXk=(1+ϵ)p+(1−ϵ)(1−p). Similarly  Eyn2=E∏kXk2=∏kEXk2=(EXk2)n  again valid for any distribution. In our case: EXk2=(1+ϵ)2p+(1−ϵ)2(1−p). In particular, var=(EXk2)n−((EXk)2n and you may carry on from there, in order to calculate limits etc... (e.g. n→∞, ϵn→λ gives a nice limit). The actual distribution of yn is in general quite complicated.

Lakisha Archer · Accepted Answer

Assuming independence and regarding the expectation:  For the sake of simplicity, let y0=1. The value of our variable, at the nth moment, is  (1+ϵ)k(1−ϵ)n−k  with probability  (nk)pk(1−p)n−k  (Where k denotes the number of multiplications by 1+ϵ.)  The expected value is, then  ∑k=0n(1+ϵ)k(1−ϵ)n−k(nk)pk(1−p)n−k=  =∑k=0n(nk)[(1+ϵ)p]k[(1−ϵ)(1−p)]n−k=  =[(1+ϵ)p+(1−ϵ)(1−p)]n=[1+2ϵp−ϵ]n  because of the binomial theorem.  Unfortunately I could not find the trick for the calculation of the expectation of the square of the same random variable...

Vasquez · Accepted Answer

Following up zoli&#039;s answer, I think E(yt2) can be found the same way, with the Binomial theorem yielding (assuming y0=1) E(yt2)=y02E((1+ϵ)2(1−ϵ)2) =[(1+ϵ)2p+(1−ϵ)2(1−p)]n =[(1−ϵ)2+4ϵp]n So the variance is σn2=[(1−ϵ)2+4ϵp]n−[1+2ϵp−ϵ]2n. I don&#039;t know if it can be simplified any further. Quick check: as ϵ→0, σn→0

I have a real valued number y_t. At each time

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